Listed below are speeds (mi/h) measured from traffic on a busy highway. This simple random sample was obtained at 3:30 P. M. on a weekday. Use the sample data to construct a 98% confidence interval estimate of the population standard deviation. 65 62 62 55 62 55 60 59 60 70 61 68 The confidence interval estiamte it __mi/h< o <__mi/h

 3.0 mi/h < σ < 8.54 mi/h

Step-by-step explanation:

Given:

Sample data:  x: 65 62 62 55 62 55 60 59 60 70 61 68

Confidence = c = 98% = 0.98

To find:

Construct a 98% confidence interval estimate of the population standard deviation.

Solution:

Compute Mean:

number of terms in data set = n = 12

Mean = Sum of all terms / number of terms

          = 65 + 62 + 62 + 55 + 62 + 55 + 60 + 59 + 60 + 70 + 61 + 68 / 12

          = 739/12

Mean  = 61.58

Compute standard deviation:

s = √∑(each term of data set - mean)/ sample size - 1

s = √∑()²/n-1

  = √( 65 - 61.58)² + (62 - 61.58)² + (62 - 61.58)² + (55 - 61.58)² + (62 - 61.58)² + (55 - 61.58)² + (60 - 61.58)² + (59 - 61.58)² + (60 - 61.58)² + (70 - 61.58)² + (61 -61.58)² + (68 - 61.58)² / 12-1

  = √(11.6964 + 0.1764 + 0.1764 + 43.2964 + 0.1764 + 43.2964 + 2.4964 + 6.6564 + 2.4964 + 70.8964 + 0.3364 + 41.2164) / 11

  = √222.9168/11

  = √20.2652

  = 4.50168

  = 4.5017

s =  4.5017

Compute critical value using chi-square table:

For row:

degree of freedom = n-1 = 12 - 1 = 11

For Column:

(1 - c) / 2 = (1 - 0.98) / 2 = 0.02/2 = 0.01

1 - (1 - c) / 2 = 1 - (1-0.98) / 2 = 1 - 0.02 / 2 = 1 - 0.01 = 0.99

= 3.053

= 24.725

Compute 98% confidence interval of standard deviation:

 = = =  

             = 0.6670 (4.5017) = 3.0026

 =  3.0026

=   =   =  

               =  1.8982 ( 4.5017) = 8.5449

 = 8.5449

3.0026 mi/h < σ < 8.5449 mi/h

2 x t divided by n

step-by-step explanation:

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: